5

I am looking for symptotic formula for $\sum_{n<x} \frac {d(n)}{\sqrt{n}}$ which doesn't use $\zeta(\frac{1}{2})$

My guess is - perhaps it is something like $A\sqrt{x}\log{x} + B \sqrt{x} + O(\log{x})$ where $A, B$ are some constants, based on these formulas:

$$\sum_{n<x} \frac {d(n)}{n} = \frac{1}{2}(\log{x})^2 + 2\gamma\log{x} + O(1)$$ $$\sum_{n<x} {d(n)} = x\log{x} + (2\gamma-1)x + O(\sqrt{x})$$

Per Huxley the error term can be improved to $$\sum_{n<x} {d(n)} = x\log{x} + (2\gamma-1)x + O(x^\theta)$$ with $ \inf \theta \le 131/416 = 0.31490384615 $

Similar questions:

Asymptotic for $\sum_{n<x} \frac {d(n)}{\sqrt{n^a}}$

Asymptotic for $\sum_{n<x} d(n)$

Shree
  • 733
  • To clarify -- are you looking for something tighter than the $2\sqrt x\log x+O(\sqrt x)$ bound given in the first question linked? – Carl Schildkraut Apr 19 '22 at 22:16
  • Do you know the classical proof of your last displayed formula using the hyperbola method? That same method should work here, probably getting an error that's $O(x^{1/4})$. – Greg Martin Apr 19 '22 at 22:20
  • 1
    what about using summation by parts using Voronoi formula $\sum_{n<x} {d(n)} = x\log{x} + (2\gamma-1)x + O(x^{1/3})$ to get a better asymptotic – Conrad Apr 19 '22 at 23:37
  • Thank you all. I will work on these ideas. – Shree Apr 20 '22 at 00:00

1 Answers1

5

Use Abel's summation formula.

If $$A(x) = \sum_{n< x} d(n) = x \log x +(2\gamma -1) x + O(\sqrt{x})$$

then $$\sum_{n< x}\frac{d(n)}{\sqrt{n}} = \frac{A(x)}{\sqrt{x}} + \int_1^x\frac{A(t)}{2\ t^{3/2}}\ dt$$

From this you should be able to get

$$\sum_{n< x}\frac{d(n)}{\sqrt{n}} = 2\sqrt{x}\log x + 4(\gamma -1)\sqrt{x}+O(\log x)$$

jjagmath
  • 22,582
  • 1
    if needed one can get a better error $O(1)$ by using known better errors in $A$ (eg Voronoi's error is $O(x^{1/3})$ and there are better ones too) – Conrad Apr 20 '22 at 00:16
  • @Conrad Just a detail: Voronoi's error is $O(x^{1/3}\log x)$, often written in a less precise way as $O(x^{1/3+ \varepsilon})$ for all $\varepsilon > 0$, but not $O(x^{1/3})$. – jjagmath Apr 21 '22 at 19:07
  • Wouldn't dispute that but there are are lower exponents than $1/3$ that are known to work though they require slightly more work – Conrad Apr 21 '22 at 20:05